Prove : $\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$ with $a+b+c=1$

Question

Claim :

Let $a\geq b\geq c\geq 0$ and $a+b+c=1$ then we have :

$$\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$$

To prove it I try the Ji chen's theorem because we have :

$$\frac{a^2+b^2+1}{a^2+c^2+1}+\frac{a^2+c^2+1}{b^2+c^2+1}+\frac{b^2+c^2+1}{a^2+b^2+1}\geq 3$$ $$\frac{a^2+b^2+1}{a^2+c^2+1}\frac{a^2+c^2+1}{b^2+c^2+1}+\frac{a^2+b^2+1}{a^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}+\frac{a^2+c^2+1}{b^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}\geq 3$$ $$\frac{a^2+b^2+1}{a^2+c^2+1}\frac{a^2+c^2+1}{b^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}=1$$

To prove the two first I use Karamata's inequality as we have .

Let $x,y,z,u,v,w>0$ such that $x\geq y \geq z$ and $u\geq v \geq w$ and : $x\geq u$, $xy\geq uv$, $xyz\geq uvw$

Then we have : $$x+y+z\geq u+v+w$$

Question :

Is there simpler ?

Thanks in advance !

Regards

Max

Answer 1

Hint:clearly this is of form $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge 3$$ which is true by am-gm